Answer:
Step-by-step explanation:
If base of variable is same, add the powers
1) x² * x * x³ * x⁵ = x²⁺¹⁺³⁺⁵ = x¹¹
2) -2n² * -n⁴ * 5n³ = (-2* -1* 5 ) *n²⁺⁴⁺³ = 10n⁹
3)-3x⁴ * 2x⁵ * - x = (-3 * 2 * -1)* x⁵⁺⁴⁺¹ = 6x¹⁰
If we expand bx to jx and kx we have:
5y^2-2y-7
5y^2-7y+5y-7 then factor...
y(5y-7)+1(5y-7)
(y+1)(5y-7)
So the other factor is:
(y+1)
Answer:
1. 15x^7y^2 + 4x^3 => x^3(15x^4y^2 + 4)
2. 15x^7y^2 + 3x => 3x(5x^6y^2 + 1)
3. 15x^7y^2 + 6xy => 3xy(5x^6y + 2)
4. 15x^7 + 10y^2 => 5(3x^7 + 2y^2)
Step-by-step explanation:
To obtain the answer to the question, first let us factorise each expression. This is illustrated below:
1. 15x^7y^2 + 4x^3
Common factor is x^3, therefore the expression is written as:
x^3(15x^4y^2 + 4)
2. 15x^7y^2 + 3x
Common factor is 3x, therefore the expression is written as:
3x(5x^6y^2 + 1)
3. 15x^7y^2 + 6xy
Common factor is 3xy, therefore the expression is written as:
3xy(5x^6y + 2)
4. 15x^7 + 10y^2
Common factor is 5, therefore the expression can be written as:
5(3x^7 + 2y^2)
Answer:
that answer is D
Step-by-step explanation:
I used pythagreum theurum a^2+b^2=c^2
then i divided square root 260 by 4 the largest perfect square factor which gives us 2 square root 65 because 4 is a perfect square that equal 2
